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A052297 Number of distinct prime factors of all composite numbers between n-th and (n+1)st primes. 13
0, 1, 2, 3, 2, 4, 2, 5, 5, 3, 6, 5, 3, 5, 6, 7, 3, 7, 6, 2, 8, 4, 8, 9, 5, 3, 6, 2, 6, 14, 5, 8, 3, 11, 3, 9, 7, 6, 8, 8, 3, 13, 2, 6, 3, 14, 15, 5, 3, 7, 9, 3, 11, 8, 9, 9, 3, 9, 6, 3, 13, 16, 7, 3, 6, 16, 8, 13, 3, 6, 9, 10, 9, 9, 6, 8, 11, 6, 12, 14, 4, 14, 2, 10, 7, 8, 11, 6, 4, 6, 16, 10, 6, 13 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

From Lei Zhou, Mar 18 2014: (Start)

This is also the number of primes such that the (n+1)-th prime (mod i-th prime) is smaller than the (n+1)-th prime (mod n-th prime) for 1 <= i < n.

Proof: We denote the n-th prime number as P_n. Suppose P_(n+1) mod P_i = k; we can write P_(n+1) = m*P_i + k. Setting l = P_(n+1) - P_n, the composite numbers between P_n and P_(n+1) will be consecutively m*P_i + C, where C = k-l+1, k-l+2, ..., k-1. If k < l, there must be a value at which C equals zero since k-1 > 0 and k-l+1 <= 0, so P_i is a factor of a composite number between P_n and P_(n+1). If k >= l, all C values are greater than zero, thus P_i cannot be a factor of a composite number between P_n and P_(n+1). (End)

LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000

EXAMPLE

n=30, p(30)=113, the next prime is 127. Between them are 13 composites: {114, 115, ..., 126}. Factorizing all and collecting prime factors, the set {2,3,5,7,11,13,17,19,23,29,31,41,59,61} is obtained, consisting of 14 primes, so a(30)=14.

MATHEMATICA

Length[Union[Flatten[Table[Transpose[FactorInteger[n]][[1]], {n, First[#]+ 1, Last[#]-1}]]]]&/@Partition[Prime[Range[100]], 2, 1] (* Harvey P. Dale, Jan 19 2012 *)

CROSSREFS

Cf. A052180, A052248, A061214, A077218.

Sequence in context: A222817 A309428 A328368 * A297162 A322587 A058973

Adjacent sequences:  A052294 A052295 A052296 * A052298 A052299 A052300

KEYWORD

nonn

AUTHOR

Labos Elemer, Feb 09 2000

STATUS

approved

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Last modified September 25 00:02 EDT 2020. Contains 337333 sequences. (Running on oeis4.)